Question

questuon in in attachment who knws give ans​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

Step-by-step explanation:

given:

$\sqrt{y+1} - \sqrt{y-1} = \sqrt{4y-1}$

square on both sides

$\sqrt{y+1} ^{2} - 2\sqrt{y+1} \sqrt{y-1} + \sqrt{y-1}^{2} = \sqrt{4y-1} ^{2} \\\\y+1+y-1- 2\sqrt{y+1} \sqrt{y-1}=4y-1\\\\2y-2\sqrt{y+1} \sqrt{y-1}=4y-1\\\\ 2\sqrt{y+1} \sqrt{y-1}=2y-4y+1\\square on both sides\\4 (y+1)(y-1)=(1-2y)^2\\4(y^2-1)=1-4y+4y^2\\4y^2-4=1-4y+4y^2\\4y=5\\y=5/4$

Hopefully this is the right answer and you have understood the method